LeetCode: Coin Change

Posted on February 26, 2018July 26, 2020 by braindenny

Coin Change

Similar Problems:

You are given coins of different denominations and a total amount of money amount. Write a function to compute the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.

Example 1:

coins = [1, 2, 5], amount = 11
return 3 (11 = 5 + 5 + 1)

Example 2:

coins = [2], amount = 3
return -1.

Note:
You may assume that you have an infinite number of each kind of coin.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

## https://code.dennyzhang.com/coin-change
class Solution(object):
    ## Basic Ideas: Dynamic Programming
    ##
    ##    dp(i) = min(dp(i), dp[i-coin[j]]+1)
    ## Complexity: Time O(n), Space O(n)
    def coinChange(self, coins, amount):
        """
        :type coins: List[int]
        :type amount: int
        :rtype: int
        """
        if amount == 0: return 0
        import sys
        coins.sort()
        dp = [0] + [sys.maxsize]*(amount)

        for i in range(1, amount+1):
            for coin in coins:
                if coin > i: break
                if dp[i-coin] != sys.maxsize:
                    dp[i] = min(dp[i], dp[i-coin]+1)

        return dp[amount] if dp[amount] != sys.maxsize else -1 

    ## Basic Ideas: BFS
    ##
    ## Complexity: Time O(n), Space O(n)
    def coinChange_v1(self, coins, amount):
        """
        :type coins: List[int]
        :type amount: int
        :rtype: int
        """
        if amount == 0: return 0
        coins.sort()
        d = {}
        queue = collections.deque()
        queue.append(0)
        level = 0
        while len(queue) != 0:
            level += 1
            for k in range(len(queue)):
                num = queue.popleft()
                for coin in coins:
                    v = coin + num
                    if v > amount: break
                    if v == amount: return level
                    # avoid duplicate
                    if v not in d:
                        d[v] = level
                        queue.append(v)
        return -1

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